Sorry but still wrong. As Franklin2 says DC (Direct Current) has the property to try to keep flowing once the circuit is interrupted. The electrical theory for Inductive circuits is very clear and a couple of long wires running parallel and separate by insulator and air is not considered as a circuit with inductive properties. Filament bulb is basically a coil but there is not any magnetic core, just vacuum or inert gas inside bulb, then it doesn't have inductive properties.

If you want have a real idea on how inductive is a circuit, then just energize it using AC current, then you will see all the inductive properties been enhanced by AC current.

Try this:

1º At the Engine Comp. Fuse Box locate the Mega fuse which provides power to head light switch. Remove it.

2º Using a 250VA transformer reducer from 127 to 12Vac, connect one secundary wire to chasis ground and the another wire to the pin at the Fuse box which goes to head light switch.

3º Turn on the head lights and then mesure current and voltage in both, high & low beam. Take a look of the head switch while you turn on - off several times just to notice if spark is bigger or smaller than DC spark

4º Disconnect everything and replace the mega fuse, do same voltage & current measures.

5º If the circuit has important inductive properties, you will notice that power (W=V*I) used by lights with AC is a lot lower than used with DC. this is because the Inductive component of the circuit will produce a total impedance which always is bigger than just the resistive component of the circuit.

If the calculations gives almost same poer used, will confirm the circuit doesn't have important Inductive properties.

 

Not having a magnetic core doesn't mean the coil isn't an inductor. Air-core inductors are used all the time in switchmode power supplies because of their lower saturating current. Coil = inductor regardless of the core type. Now, it is true that air core inductors can't have as high of an inductance per package size, but they are still inductors. Saying a light bulb does not act as in inductor is not correct at all.

I said before that voltage across an inductor is proportional to the rate of change of current through it. That means V = L*(di/dt) where L is the inductance and di/dt is the derivative of current. You can find that in any textbook. If you try and force current to zero from several amps, that means the derivative shoots up incredibly fast, and V becomes very very large for a short amount of time. That means a huge voltage. That inductor is parallel across the open switch, that's why you see a spark across the open contacts. The inductor doesn't have to be very big at all. It takes a lot of voltage to spark through air, even when the contacts aren't very far apart. The only place you're going to get a voltage that big is from an inductor trying to discharge open-circuit.

I'm sorry, but in the 4 years I spent getting my bachelor's degree in electrical engineering, no one ever mentioned a magic property of direct current trying to keep moving forward in a purely resistive circuit that has no indutance. Where does the extra energy come from then, if there is zero inductance? There is no way around Ohm's law. If someone can show me a reputable source explaining how DC in a resistive circuit wants to keep flowing when a voltage source is taken away, then I will take it into consideration and appreciate the help. We're all here to learn anyways.

 

I appreciate the above comment, I am here to learn too.

While you can't believe everything you read on the internet, or a book for that matter, here are some links on the DC subject.

http://www.safetydata.com/switches.htm

Here's a site that describes the different type loads, and talks about the high inrush current of a bulb.

http://www.carlingtech.com/products/...hes_amp-rating

Here's a site from a guy who plays with electric cars. I was building one too, and that's were I learned about the differences between AC and DC circuits, and the problem of breaking a DC circuit.
http://www.mrsharkey.com/busbarn/electrics/chapt9.htm

 

Interesting links and thank you for sharing those.

There is no doubt that arcing with DC switches is a problem. However, I did not find anything saying that DC wants to move after interruption simply because and only because it is DC. My stance - and this seemed to be what I gathered from these links too - is that it is the loads that lead to the arcing. DC simply harbors the arcing more violently than AC. I think this is a consensus between all of us, including the links. The first link, in particular, mentioned breaker points can suffer from arcing. Points are connected to the coil - the inductance from the coil is what leads to the arcing. If the coil was simply a resistor and had no inductive properties at all, there wouldn't be any arcing.

I think all of those links are good sources and I agree with what they say. I'm not insisting that DC circuits don't suffer from arcing more than they do in AC circuits; I think we are all pretty clear on that. DC indeed wants to keep moving forward after you throw the headlight switch - but this is because there is inductance in the circuit and the headlight circuit is not purely resistive. My stance is that this tendency to keep flowing with the switch open, is not caused by a magic property of DC and DC only - it is caused by the inductance of the bulbs. javi2001's post claimed that the bulbs cannot act as inductors, but I do not agree and I can prove it mathematically. I can explain mathematically why the arc occurs. My stance is that such an arc (which takes a high voltage) can be caused by nothing else but inductance.

If you run DC through a pure resistor and open the switch, it cannot arc. There is no energy storage device in this case that will keep current trying to flow. There is no inductance in this imaginary circuit, so L = 0. V = L*di/dt, so it will be zero, and therefore there will be no voltage across this imaginary switch, and hence no arcing. However, you are right that fuses have a rating for DC and AC. But this is because people usually hook fuses up to circuits that have loads with inductance - heaters, motors, solenoids, and light bulbs. Anything with a coil of wire. And when you have such an inductive load in the mix, the DC circuit is in danger of arcing. That is why for a given fuse, it is safer for less DC amperage than it would be for AC amperage, because DC will arc more for a given inductive load than it will for AC. No doubt about that. My only point is that the arcing is from the inductance, not from a property of DC specifically. The arcing can show up for AC too, just not as harmfully, hence a more forgiving rating for the same fuse in an AC application.

I think the DC vs. AC ratings of fuses cause confusion as to why DC arcs like it does. The fact that a fuse is rated for less DC amperage than AC doesn't mean that DC has an innate physical property to be "hard to stop" regardless of load type. DC simply arcs more with loads that we commonly hook fuses to, than it does for AC. It is the inductive loads that cause the arcing.

 

Ok let's see...

The ability of a DC current to try to keep flowing is not Magic.

Do a simple test for teens lab...

Get a pencil, remove the wood from the graphite and cut a piece of graphite about 1 inch long.

This will be what almost any electrician consider a "pure" resistor, no inductive properties, no capacitive properties...

Get 2 wires #14 AWG and 4 inches long each, connect to our graphite resistor and go to your truck battery, after connect and disconnect several times you will see our "pure" resistive circuit produce sparks !!!!

Doing same test with 24 Vdc will produce bigger sparks !!!

You should go back to the beginning of your electrical studies and remember some basics theories which in my opinion can explain this interesting behavior...

1º Electrical current is basically a flow of electrons through a conductor in a closed circuit

2º Those electrons has mass, very small but they has...

3º Electrons moves through a conductor at the Light speed (Around 300.000 Km/h)

4º Just 1 Ampere/hour are millions of electrons...

5º A basic physics theory says any mass moving has Inertia, (Potential energy); proportional to the mass and the Speed.

So once the electrons current are flowing they has inertia... and try to keep moving....

Going back to our initial post, in my opinion; the head lights plug is melted by deficient electrical contact between connector and switch. The spark produced in the normal operation of the switch, doesn't have enough energy to heat the whole assembly until melt it.

This will be my last post out of the original topic "Head light switch plug melted! "

 

If your approach to this discussion is to take it to a personal level, then I'm not replying anymore either. I explained in detail why an inductive load makes a switch arc when you switch it off; take it or leave it. The sparks from the test you are talking about are not in the voltage range it takes for current to arc across switch contacts like this. I stated many times that the same thing occurs with AC circuits too and you can see for yourself when you switch a light off in a room without windows. I'm sure the OP got what he needed from this thread, and that's what's important. Good day to you sir.

 

Doing this should be a big benefit.

http://www.northwestvintagebroncos.c...headlight1.htm

 

That's the beauty of these discussion boards. It's not like you are having a discussion with someone in the same room or in a group, and feel you need to yell, or say something over and over to be heard. It's all here in print, every opinion is here for everybody to read, and they can draw their own conclusions after reading the whole thread, or they can get interested in the subject, do some research, and state their own opinion in the thread, and have their opinion printed in the thread like everyone else's.

 

Well, mine is a Diesel with batteries on both sides, and I think I will just sever the wiring at a point just before where it splits off to go to both lights and use the wires going to the lights right there to activate the relays.

I do need to make a new high beam switch and relocate it as the one on the floor is gone and the spot where it would mount is rusted through.

Another 1/2 assed repair made by the previous owner, or the one who sold it to him.